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我已经在网上搜索了几个小时,在整个“外部 js 文件”-jungle 中找不到答案。我希望你们能帮忙!

简而言之:我的外部 javascript 文件似乎没有得到我在 main.php 文件中定义的变量。

  1. main.php我定义 php 变量并将它们“转换”为 javascript 变量

    <head>...
    <script type="text/javascript">
    var phpmain_img = <?php echo json_encode($main_img); ?>;
    var phpvar1_large = <?php echo json_encode($var1_large); ?>;
    var phpvar2_large = <?php echo json_encode($var2_large); ?>;
    var phpvar3_large = <?php echo json_encode($var3_large); ?>;
    var phpvar4_large = <?php echo json_encode($var4_large); ?>;
    </script>
    ...
    <script language="javascript" type="text/javascript" src="/wshop/ext.js"></script>
    </head>
    
  2. 在我的ext.js文件中,我想处理这些变量。在 ext.js 文件中,我定义了一个将在主 PHP 中使用的函数 swapImage():

            var imgArray = new Array(
            phpmain_img,
            phpvar1_large,
            phpvar2_large,
            phpvar3_large
            );
    
        function swapImage(imgID) {
            var theImage = document.getElementById('theImage');
            var newImg;
            newImg = imgArray[imgID];
            theImage.src = newImg;
        }
    
        function preloadImages() {      
            for(var i = 0; i < imgArray.length; i++) {
                var tmpImg = new Image;
                tmpImg.src = imgArray[i];
            }
        }
    
  3. 结果:main.php中的 swapImage() ... 不起作用

    <div id="image">
        <img id="theImage" src="<?=$main_img; ?>" alt="" />
    </div>
    
    <div id="thumbs">
    <?php
    
            echo "<img src=\"<$main_img_small\" alt=\"\" onmouseover=\"swapImage(0);\">";
            echo "<img src=\"$var1_small\" alt=\"\" onmouseover=\"swapImage(1);\">";
            echo "<img src=\"$var2_small\" alt=\"\" onmouseover=\"swapImage(2);\">";
            echo "<img src=\"$var3_small\" alt=\"\" onmouseover=\"swapImage(3);\">";
    
            ?>
    
    
        <br />
    </div>
    

任何帮助是极大的赞赏!

更新:

我没有收到特定错误,swapImage 函数在鼠标悬停时不起作用。但是,我尝试使用 eg 输出变量,document.write(phpimg_main)但没有出现任何内容,这让我相信移交变量有问题...

这是源代码浏览器输出

<html>
    <head>
        <link href="../demo.css" rel="stylesheet" type="text/css" />
        <style type="text/css">
            ....
</style>

        <script type="text/javascript">
        var phpmain_img = {"0":"http:\/\/path\/to\/main\/image.jpg"};
        var phpvar1_large = {"0":"http:\/\/path\/to\/image1.jpg"};
        var phpvar2_large = {"0":"http:\/\/path\/to\/image2.jpg"};
        var phpvar3_large = null;
        var phpvar4_large = null;
        </script>   

        <script language="javascript" type="text/javascript" src="/wshop/ext.js"></script>
</head>




<body onload="preloadImages()">

    <div id="image">
        <img id="theImage" src="http://path-to-main-image.jpg" alt="" />
    </div>

    <div id="thumbs">
    <img src="http://path-to-main-image.jpg" alt="" onmouseover="swapImage(0);"><img src="http://path-to-image1.jpg" alt="" onmouseover="swapImage(1);"><img src="http://path-to-image2.jpg" alt="" onmouseover="swapImage(2);">

        <br />
    </div>



</body>

`

更新 2:

感谢您的输入和回答!当然,你是对的,我需要一个字符串而不是一个对象,所以编码是一个很好的提示。

然而,问题仍然没有用 [0] 解决。即使我像下面这样对其进行硬编码,第二个 javascript 块(我之前尝试将其外包为外部 js 文件)也不会获得第一个 javascript 块中定义的变量。

<script type="text/javascript">
        var phpmain_img = "http://www.abc.de/path-img_main.jpg";
        var phpvar1_large = "http://www.abc.de/path-img1.jpg";
        var phpvar2_large = "http://www.abc.de/path-img2.jpg";
        var phpvar3_large = "http://www.abc.de/path-img3.jpg";
        var phpvar4_large = "http://www.abc.de/path-img4.jpg";
        </script>

<script language="javascript" type="text/javascript">


                   var imgArray = new Array(
                phpmain_img,
                phpvar1_large,
                phpvar2_large,
                phpvar3_large,
                phpvar4_large
            );



            function swapImage(imgID) {
                var theImage = document.getElementById('theImage');
                var newImg;
                newImg = imgArray[imgID];
                theImage.src = newImg;
            }

            function preloadImages() {      
                for(var i = 0; i < imgArray.length; i++) {
                    var tmpImg = new Image;
                    tmpImg.src = imgArray[i];
                }
            }



            </script>

浏览器源视图输出:

<script type="text/javascript">
        var phpmain_img = "http://www.abc.de/path-img_main.jpg";
        var phpvar1_large = "http://www.abc.de/path-img1.jpg";
        var phpvar2_large = "http://www.abc.de/path-img2.jpg";
        var phpvar3_large = "http://www.abc.de/path-img3.jpg";
        var phpvar4_large = "http://www.abc.de/path-img4.jpg";
        </script>

        <script language="javascript" type="text/javascript"> //this i actually wanted to outsource into an external js-file

        var imgArray = new Array(
            phpmain_img,
            phpvar1_large,
            phpvar2_large,
            phpvar3_large,
            phpvar4_large
        );



        function swapImage(imgID) {
            var theImage = document.getElementById('theImage');
            var newImg;
            newImg = imgArray[imgID];
            theImage.src = newImg;
        }

        function preloadImages() {      
            for(var i = 0; i < imgArray.length; i++) {
                var tmpImg = new Image;
                tmpImg.src = imgArray[i];
            }
        }



        </script>

对不起,这里的话题很长。我希望你能跟随!我正在努力学习!

4

1 回答 1

1

虽然我不是 PHP 专家,但你可能是说

var phpmain_img = "<?php echo urlencode($main_img); ?>";

代替

var phpmain_img = <?php echo json_encode($main_img); ?>;

如果您查看呈现的标记,很明显图像变量不包含您的预期:

var phpmain_img = {"0":"http:\/\/path\/to\/main\/image.jpg"};

它不是一个字符串,而是一个对象。

于 2012-11-22T22:56:41.290 回答